Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{t^2 - 10t + 16}{7t - 56} \div \dfrac{6t - 12}{t + 1} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{t^2 - 10t + 16}{7t - 56} \times \dfrac{t + 1}{6t - 12} $ First factor the quadratic. $y = \dfrac{(t - 2)(t - 8)}{7t - 56} \times \dfrac{t + 1}{6t - 12} $ Then factor out any other terms. $y = \dfrac{(t - 2)(t - 8)}{7(t - 8)} \times \dfrac{t + 1}{6(t - 2)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (t - 2)(t - 8) \times (t + 1) } { 7(t - 8) \times 6(t - 2) } $ $y = \dfrac{ (t - 2)(t - 8)(t + 1)}{ 42(t - 8)(t - 2)} $ Notice that $(t - 8)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ \cancel{(t - 2)}(t - 8)(t + 1)}{ 42(t - 8)\cancel{(t - 2)}} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $y = \dfrac{ \cancel{(t - 2)}\cancel{(t - 8)}(t + 1)}{ 42\cancel{(t - 8)}\cancel{(t - 2)}} $ We are dividing by $t - 8$ , so $t - 8 \neq 0$ Therefore, $t \neq 8$ $y = \dfrac{t + 1}{42} ; \space t \neq 2 ; \space t \neq 8 $